ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A finite-grid L-curve selects an interior parameter

Example

Take $A = I_{2} \in R^{2 \times 2}$ , $b = e_{1}$ , and the strictly increasing grid $1 / 9 < 1 < 9$ . Apply the discrete maximum-curvature L-curve selector to this input. The Tikhonov solution at $λ$ is $x_{λ} = (1 + λ)^{- 1} e_{1}$ , so all three residual and solution norms are positive and the selector returns the middle grid value $λ = 1$ . This finite-grid data-driven choice is made without a noise-level input, and it carries no universal optimality claim. The matrix is finite-dimensional, so the discontinuity hypothesis of the Bakushinskii veto is absent.

Example references

Discrete maximum-curvature L-curve selector

Hansen, Introduction to Inverse Problems, Chapter 5, slides 10–14

Facts & Assumptions

Given: The real matrix $A = I_{2}$ , the datum $b = e_{1}$ , and the grid $(λ_{1}, λ_{2}, λ_{3}) = (1 / 9, 1, 9)$ .

[F1]

Discrete maximum-curvature L-curve parameter selection: At each grid value the selector computes $x_{j} = (A^{T} A + λ_{j} I)^{- 1} A^{T} b$ , its residual and solution norms, and discards zero residuals or solution norms. For consecutive retained log-log points $p_{0}, p_{1}, p_{2}$ , it sets $u = p_{1} - p_{0}$ and $v = p_{2} - p_{1}$ , then defines $κ = 2 ∣ \det (u, v) ∣ / (∥ u ∥_{2} ∥ v ∥_{2} ∥ u + v ∥_{2})$ when the denominator is positive and $0$ otherwise. It assigns this curvature to the middle index and returns a parameter with largest curvature, breaking ties by smallest original index.

Verification

1.1givenF1algebra

Since $A = I_{2}$ , the formula in [F1] gives $x_{λ} = (1 + λ)^{- 1} e_{1}$ , residual norm $r_{λ} = ∥ A x_{λ} - b ∥_{2} = λ / (1 + λ)$ , and solution norm $s_{λ} = ∥ x_{λ} ∥_{2} = 1 / (1 + λ)$ . At the three grid values, $(r_{λ}, s_{λ})$ equals $(1 / 10, 9 / 10)$ , $(1 / 2, 1 / 2)$ , and $(9 / 10, 1 / 10)$ , respectively. All retained points are valid because both coordinates are positive.

2.1step 1.1F1algebra

Their log-log points are $p_{0} = (- \log 10, \log (9 / 10))$ , $p_{1} = (- \log 2, - \log 2)$ , and $p_{2} = (\log (9 / 10), - \log 10)$ . Thus $u = p_{1} - p_{0} = (\log 5, - \log (9 / 5))$ and $v = p_{2} - p_{1} = (\log (9 / 5), - \log 5)$ . Their determinant is $- \log^{2} 5 + \log^{2} (9 / 5) \neq 0$ because $5 > 9 / 5 > 1$ ; moreover $∥ u ∥_{2} = ∥ v ∥_{2} > 0$ and $u + v = (\log 9, - \log 9) \neq 0$ . The algorithm's curvature is therefore well-defined and positive.

3.1givenstep 2.1F1algebra∎

There are exactly three retained grid indices, so the only consecutive triple is $(1, 2, 3)$ and its curvature is assigned to index $2$ . This is the only candidate for the maximum, so the selector returns $λ_{2} = 1$ ; no tie-break is needed. The computation supplies no claim about any grid outside this input.

Depends on

Discrete maximum-curvature L-curve parameter selection

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Per Christian Hansen, Introduction to Inverse Problems, Chapter 5: Regularization Parameter Choice (standard reference, not scraped)